Question

求解一道分式方程。~

2/3x=x^2/12+3/x^2+4/x
还有一道 x+1/x-1=7/2

Answer 1

2/3x=x^2/12+3/x^2+4/x 

两边同时乘以12x^2

x^2-8x+12=0解得x=2或6

Answer 2

2/3x=x^2/12+3/x^2+4/x

问题补充：还有一道 x+1/x-1=7/2

(x^2+1)/x=9/2 x=（根号65+9）/4或=（-根号65+9）/4

Answer 3

2x/3=x²/12+3/x²+4/x
2x/3*12x²=x²/12*12x²+3/x²*12x²+4/x*12x²
8x³=x^4+36+48x
x^4-8x³+48x+36=0
(x²-2x-6)(x²-6x-6)=0
x=(2±√28)/2或x=(6±√60)/2
x=(2±2√7)/2或x=(6±2√15)/2
x=(1±√7)或x=(3±√15)

(x+1)/(x-1)=7/2
2(x+1)=7(x-1)
2x+2=7x-7
-5x=-9
x=9/5